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From the unbalanced reaction: B2H6 + O2 ---> HBO2 + H2O

How many grams of O2 (32g/mol) will be needed to burn 36.1 g of B2H6 (Molar mass = 27.67g/mol)? ______g


Include the correct number of significant figures in your final answer

User Betaveros
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1 Answer

3 votes

Answer: 125 g

Step-by-step explanation:

To calculate the moles :


\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}


\text{Moles of} B_2H_6=(36.1g)/(17)=1.30moles

The balanced reaction is:


B_2H_6+3O_2\rightarrow 2HBO_2+2H_2O

According to stoichiometry :

1 mole of
B_2H_6 require = 3 moles of
O_2

Thus 1.30 moles of
B_2H_6 will require=
(3)/(1)* 1.30=3.90moles of
O_2

Mass of
O_2=moles* {\text {Molar mass}}=3.90moles* 32g/mol=125g

Thus 125 g of
O_2 will be needed to burn 36.1 g of
B_2H_6

User Configurator
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